What is the percent composition of copper 2 chloride?

Answer 1

Formula: CuO

Atomic weights:

Cu- 63.5 g

O- 16.0 g

Add together (molecular mass): 79.5 g

Take the weight of copper and divide it by the total molecular mass.

63.5 / 79.5 = 79.2% Cu

There is 79.2% of copper in copper II oxide

Answer 2

Since you know how much Copper weighs and how much Chloride weighs...

Copper=63.55

Chloride=35.45

So, you add 63.55 (Cu) to 35.45 (Cl) and you get 99.00 (<-- you want 4 sig. fig.'s since you have 4 sig. fig.'s for Cu and Cl.)

From here, you get the mass of Cu and u divide by the total mass of Cu(II)Cl.

63.55/99.00=.64191919192 (<-- since we are doing % composition, we multiply by 100)

. 64191919192 x 100 = 64.19% (<-- four sig. fig.'s again DON'T FORGET... VERY CRUCIAL)

Next, we fine Chloride...

35.45/99.00=.358080808080 (<-- since we are doing % composition, we multiply by 100)

.358080808080 x 100 = 35.80% (<-- four sig. fig.'s again DON'T FORGET... VERY CRUCIAL)

In the end, we get Copper is 64.19% and Chloride is 35.80%

P.S. in case you want to check, add the two percentage's up... they should equal 100 or at least something VERY close...

In this case, we get 99.99% (<-- prob. rounding error but pretty close)

Answer 3

Cu = 63.5 g/moleO = 16 g/mole

Copper(II) oxide = CuO = 63.5 + 16 = 79.5 g/mole

%Cu = 63.5/79.5 (x100) = 79.9%

NOTE: If you meant copper(I) oxide, it would be Cu2O and the calculation can be done the same way, but the % Cu will be greater (88.8%)

Answer 4

Copper(I) sulfide - Cu2S - has 79,75 % copper.